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stability13

# stability13 - s 1 row K> 0 or –33< K< –12 for the...

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Stability Q UESTION 13 For the system shown in Figure 13 with ( 29 ( 29 ( 29 ( 29 ( 29 2 2 1 4 1 K s G s s s s + = + + - , create the Routh table and determine for what range of K the closed–loop system is stable. + - R(s) G(s) Y(s) E(s) Figure 13 The closed–loop transfer function is ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 4 3 2 2 1 3 3 3 2 4 G s K s T s G s s s s K s K + = = + + - + + + - . The closed–loop characteristic polynomial is ( 29 ( 29 4 3 2 3 3 3 2 4 s s s K s K + - + + + - . The Routh table is ( 29 ( 29 4 3 2 1 0 1 3 2 4 3 3 0 12 2 4 0 3 33 0 0 12 2 4 0 0 s K s K K s K K K s K s K - - + - + - + + - . From the s 2 row, K < –12 for the closed–loop system to be stable. From the

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Unformatted text preview: s 1 row, K > 0 or –33 < K < –12 for the closed–loop system to be stable. From the s row, K > 2 for the closed–loop system to be stable. Since these conditions cannot be met simultaneously, the closed–loop system is unstable for all values of K . ME 279 – Automatic Control of Mechanical Systems Professor Robert G. Landers 2...
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stability13 - s 1 row K> 0 or –33< K< –12 for the...

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